Hi !
Looking again at the schematic of this great pedal, I was wondering about the role of those transistors : how are those polarised ? It looks like it is the positive part of the input signal (rectified by D2/D3) that does it, but why did they do such a thing ? That's confusing...
http://www.tonepad.com/getFile.asp?id=73
Best regards.
Eric
Q2 and Q3 serve as gates. If the input signal (as detected by the rectifier circuit, partly formed via the diodes) is strong enough to generate triggering of the flip-flop (and a stable sub-octave), then you are "allowed" to hear the output. If the input signal amplitude falls low enough, then Q2/Q3 don't turn on.
R15 and C8 provided a reasonably fast decay of the rectified signal, so that you get clean gating with minimum sputter. Keep in mind that there are TWO potential sources of "sputter" in the circuit: the flip-flop itself, and the rectifier/gate combo. In principle, you can play with the time constants of the rectified and gating action via R15/C8.
You will note that the values of R8/R5 in the first op-amp stage are to bring the signal level high enough to provide a usable rectifier signal output.
Thanx a lot Mark !
Best regards.
Eric